Question

given an = 4, d = 2, Sn = –14, find n and a.

Answer 1

dear student..
an= a+(n-1)d=4
or, a+(n-1)2=4
or, a+2n-2=4
or, a+2n=4+2
or, a+2n=6
or, a=6-2n
=n/2[2a+(n-1)d]=-14
or, n/2[2a+(n-1)2]=-14
or, an+n(n-1)=-14
or, (6-2n)n+n²-n=-14
or, 6n-2n²+n²-n=-14
or, -n²+5n=-14
or, n²-5n=14
or, n²-5n-14=0
or, n²-7n+2n-14=0
or, n(n-7)+2(n-7)=0
or, (n-7)(n+2)=0
either, n-7=0
or, n=7
or, n+2=0
or, n=-2


∵, n can not be negative ;
∴, n=7
∴, a=6-(2×7)
=6-14
=-8
∴, n=7 and a=-8 Ans.

Answer 2

A.P. Sequence. .

d = 2.. Sn = -14.

a_n = nth term= 4.
4= a1 + (n-1)d = a1+2n-2.
a1+2n = 6. Or a1 = 6-2n.
Sn = [a1 + a_n] * n/2
So: -14 = (a1 + 4)* n/2
5n-n^2+14=0
-(n-7)(n+2) = 0
n = 7 .
a1 = 6-2n= -8.